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Need help with analyzing this combination circuity

  1. Feb 27, 2009 #1
    1. The problem statement, all variables and given/known data
    the circuit
    http://img15.imageshack.us/my.php?image=38052495.gif
    http://img15.imageshack.us/my.php?image=38052495.gif
    the chart
    http://img3.imageshack.us/my.php?image=chart.gif
    http://img3.imageshack.us/my.php?image=chart.gif
    I need to fill out the chart, I have shown whats given and what I think I have solved correctly. I need help with figuring out the rest!


    2) determine the current through each resistor
    http://img3.imageshack.us/my.php?image=current.gif
    no idea on how to approach this

    2. Relevant equations



    3. The attempt at a solution
     
    Last edited: Feb 27, 2009
  2. jcsd
  3. Feb 27, 2009 #2
    For the first 2 images:
    I would suggest simplifying the circuit into its Voltage source and a single Req. You can use this information to determine the source current. Then expand the circuit back out to its original form, and determine the voltage drops across each element or element combination. You will then have more than enough information to complete your table by just applying Ohm's Law where needed.
     
  4. Feb 27, 2009 #3
  5. Feb 27, 2009 #4
    nope
    for the first one how can i get req? I dont know any of the resisort values
    and is the total resistance for 2, 4.5?
     
  6. Feb 27, 2009 #5
    Oh OK, I didn't see that at first. Forget Req. :biggrin:

    Your R1 is correct, but your R2 isn't. I think you had the right idea but just messed up the math.

    Think about the given currents in R1 and R2 and how that relates to the source current. With that, you can know what the currents are through all the elements, and that may help you to figure out many of the values. Also, remember that the total sum of the voltage drops/increases in the loop is equal to zero.
     
  7. Feb 27, 2009 #6
    No, total resistance for 2) isn't 4.5 ohms. I would suggest redrawing the circuit so it resembles the one in the first part. The connections will be much clearer.
     
  8. Feb 27, 2009 #7
    2) would it be 3 for total resistance?
    1)
    what about my r1 is correct and what about my r2 isnt?

    would 5=current for r3?
     
  9. Feb 27, 2009 #8
    You are given the current through both resistors, R1 and R2. You are also given the voltage across R1. Notice that R1 and R2 are in parallel, which means that the voltage across R1 is also the voltage across R2, which you had correct. So you have the currents through both of the resistors (given) and the voltages across the two resistors (the same).

    V=IR ; R = V/I

    Your resistances for R1 and R2 should not be equal. You have them listed in your table as being equal.

    EDIT: Yes, current through R3 would be 5 A. Notice that this is the source current.
    No. Did you try redrawing the circuit so it resembles the circuit in 1)?
     
  10. Feb 27, 2009 #9
    got it, it should be r1=2 and r2=3, so is r5's current 4? and r6's current 5? and r3's v=15?
    2) isnt i3 in series with the battery, and i1 and i2 are in series with each otehr and they are parallel with i3?
     
    Last edited: Feb 27, 2009
  11. Feb 27, 2009 #10
    Yes to all, you are on the right track.

    Yes, R3 is in series with the battery. R1 and R2 are not in series with each other though, they are in parallel. R3 is then in series with the combination R1||R2
     
  12. Feb 28, 2009 #11
    2) so req=10? if thats right, what would I do next?

    1) r5's resistance=2, r4's voltage = 8,r4's resistance=8,? how would i figure out I and R for the battery and how would i figure out V and R for r6?
     
  13. Mar 1, 2009 #12
    Yes. Solve for the source current, and then the currents in the parallel part.

    Correct.

    Assuming the source (battery) is an ideal source, it has no internal resistance. The current through the battery is the source current you figured out earlier.

    Think about it like this. You have the voltage developed by the source (battery). You also have the voltage drops across all the resistive elements in the loop except for one, VR6. Apply KVL and you should be able to easily get the voltage drop across R6.
     
  14. Mar 1, 2009 #13
    thanks man, I really owe you a lot, just one more question, Im going to make a new thread.
     
  15. Mar 1, 2009 #14
    No problem, glad to help.
     
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